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Chapter 9 Conic Sections and Analytic Geometry The Ellipse

Chapter 9 Conic Sections and Analytic Geometry The Ellipse. P. P. F 1. F 2. Definition of an Ellipse.

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Chapter 9 Conic Sections and Analytic Geometry The Ellipse

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  1. Chapter 9Conic Sections and Analytic GeometryThe Ellipse

  2. P P F1 F2 Definition of an Ellipse • An ellipse is the set of all points in a plane the sum of whose distances from two fixed points, is constant. These two fixed points are called the foci. The midpoint of the segment connecting the foci is the center of the ellipse.

  3. The standard form of the equation of an ellipse with center at the origin, and major and minor axes of lengths 2a and 2b (where a and b are positive, and a2 > b2) is or The vertices are on the major axis, a units form the center. The foci are on the major axis, c units form the center. For both equations, b2 = a2 – c2. (0, a) (0, b) (0, c) (a, 0) (-a, 0) (0, 0) (0, 0) (-b, 0) (b, 0) (c, 0) (-c, 0) (0, -c) (0, -b) (0, -a) Standard Forms of the Equations of an Ellipse

  4. b2 = 16. This is the smaller of the two numbers in the denominator. a2 = 25. This is the larger of the two numbers in the denominator. Text Example Graph and locate the foci: 25x2 + 16y2 = 400. Solution We begin by expressing the equation in standard form. Because we want 1 on the right side, we divide both sides by 400. The equation is the standard form of an ellipse’s equation with a2 = 25 and b2 = 16. Because the denominator of the y2 term is greater than the denominator of the x2 term, the major axis is vertical.

  5. Solution Based on the standard form of the equation, we know the vertices are (0, -a) and (0, a). Because a2 =25, a = 5. Thus, the vertices are (0, -5) and (0, 5). (0, 5) (0, 3) (0, 0) (-4, 0) (4, 0) (0, -3) (0, -5) Text Example cont. Now let us find the endpoints of the horizontal minor axis. According to the standard form of the equation, these endpoints are (-b, 0) and (b, 0). Because b2 = 16, b = 4. Thus, the endpoints are (-4, 0) and (4, 0). Finally, we find the foci, which are located at (0, -c) and (0, c). Because b2 = a2 – c2, a2 =25, and b2 = 16, we can find c as follows: c2 = a2 – b2 = 25 – 16 = 9. Because c2 = 9, c = 3. The foci, (0, -c) and (0, c), are located at (0, -3) and (0, 3). Sketch the graph shown by locating the endpoints on the major and minor axes.

  6. Equation Center Major Axis Foci Vertices a2 > b2 and b2 = a2 – c2 (h, k) Parallel to the x-axis, horizontal (h – c, k) (h + c, k) (h – a, k) (h + a, k) b2 > a2 and a2 = b2 – c2 (h, k) Parallel to the y-axis, vertical (h, k – c) (h, k + c) (h, k – a) (h, k + a) y Major axis Focus (h + c, k) Vertex (h + a, k) Focus (h – c, k) Focus (h + c, k) (h, k) Major axis (h, k) Vertex (h – a, k) Vertex (h + a, k) x x Focus (h – c, k) Vertex (h + a, k) Standard Forms of Equations of Ellipses Centered at (h,k)

  7. Solution In order to graph the ellipse, we need to know its center (h, k). In the standards forms of centered at (h, k), h is the number subtracted from x and k is the number subtracted from y. This is (y – k)2 with h = -2. This is (x – h)2 with h = 1. We see that h = 1 and k = -2. Thus, the center of the ellipse, (h, k) is (1, -2). We can graph the ellipse by locating endpoints on the major and minor axes. To do this, we must identify a2 and b2. b2 = 4. This is the smaller of the two numbers in the denominator. a2 = 9. This is the larger of the two numbers in the denominator. Text Example Graph: Where are the foci located?

  8. Center Vertices Endpoints of Minor Axis 5 4 (1, -2) (1, -2 + 3) = (1, 1) (1 + 2, -2) = (3, -2) 3 2 (1, -2 - 3) = (1, 1) (1 - 2, -2) = (3, -2) (1, 1) 1 -5 -4 -3 -2 -1 1 2 3 4 5 -1 (3, -2) -2 (-1, -2) (1, -2) -3 -4 -5 (1, -5) Text Example cont. Solution The larger number is under the expression involving y. This means that the major axis is vertical and parallel to the y-axis. Because a2 = 9, a = 3 and the vertices lie three units above and below the center. Also, because b2 = 4, b = 2 and the endpoints of the minor axis lie two units to the right and left of the center. We categorize these observations as follows: With b2 = a2 – c2, we have 4 = 9 – c2, and c2 = 5. So the foci are located 5 units above and below the center, at (1,-2+ 5) and (1, -2– 5 ).

  9. Example • Find the standard form of the equation of the ellipse centered at the origin with Foci (0,-3),(0,3) and vertices (0,-5), (0,5) Solution: a = 5 and c = 3

  10. The Hyperbola

  11. y Transverse axis Vertex Vertex x Focus Focus Center Definition of a Hyperbola • A hyperbola is the set of points in a plane the difference whose distances from two fixed points (called foci) is constant

  12. y y (0, c) Transverse axis (-a, 0) Transverse axis (a, 0) (0, a) x x (-c, 0) (c, 0) (0, -a) (0, -c) Standard Forms of the Equations of a Hyperbola • The standard form of the equation of the hyperbola with center at the origin is • The left equation has a transverse axis that lies on the x-axis shown in the graph on the left. The right equation’s transverse axis lies on the y-axis. The vertices are a units from the center and the foci are c units from the center. For both equations, b2 = c2 – a2.

  13. Text Example Find the standard form of the equation of a hyperbola with foci at (0, -3) and (0, 3) and vertices (0, -2) and (0, 2). Solution Because the foci are located at (0, -3) and (0, 3), on the y-axis, the transverse axis lies on the y-axis. The center of the hyperbola is midway between the foci, located at (0, 0). Thus, the form of the equation is We need to determine the values for a2 and b2. The distance from the center (0, 0) to either vertex, (0, -2) or (0, 2) is 2, so a = 2.

  14. 5 4 3 2 1 -5 -4 -3 -2 -1 1 2 3 4 5 -1 -2 -3 -4 -5 Text Example cont. Find the standard form of the equation of a hyperbola with foci at (0, -3) and (0, 3) and vertices (0, -2) and (0, 2). Solution We must still find b2. The distance from the center, (0, 0), to either focus, (0, -3) or (0, 3) is 3. Thus, c = 3. Because b2 = c2 – a2, we have b2 = c2 – a2 = 9 – 4 = 5. Substituting 5 for b2 in the last equation gives us the standard form of the hyperbola’s equation. The equation is

  15. The Asymptotes of a Hyperbola Centered at the Origin • The hyperbola has a horizontal transverse axis and two asymptotes • The hyperbola has a vertical transverse axis and two asymptotes

  16. Graphing Hyperbolas • Locate the vertices. • Use dashed lines to draw the rectangle centered at the origin with sides parallel to the axes, crossing one axis at +a and the other at +b. • Use dashed lines to draw the diagonals of this rectangle and extend them to obtain the asymptotes. • Draw the two branches of the hyperbola by starting at each vertex and approaching the asymptotes.

  17. Example • Graph: Solution: • Center (3,1)a=2, b=5c2= 4+25= 29y= ±5/2(x-3)+1

  18. Example cont.

  19. Equation Center Transverse Axis Foci Vertices b2 = c2 – a2 (h, k) Parallel to the x-axis, horizontal (h – c, k) (h + c, k) (h – a, k) (h + a, k) b2 = c2 – a2 (h, k) Parallel to the y-axis, vertical (h, k – c) (h, k + c) (h, k – a) (h, k + a) y Vertex (h + a, k) Focus (h – c, k) Focus (h + c, k) (h, k) Focus (h + c, k) (h, k) Vertex (h + a, k) Vertex (h – a, k) Vertex (h + a, k) x x Focus (h – c, k) Standard Forms of Equations of Hyperbolas Centered at (h,k)

  20. Text Example Graph: Where are the foci located? Solution In order to graph the hyperbola, we need to know its center (h, k). In the standard forms of the equations centered at (h, k), h is the number subtracted from x and k is the number subtracted from y. We see that h = 2 and k = 3. Thus, the center of the hyperbola, (h, k), is (2, 3). We can graph the hyperbola by using vertices, asymptotes, and our four-step graphing procedure.

  21. 8 7 6 5 a = 4 a = 4 4 3 (-2, 3) (6, 3) (2, 3) 2 1 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8 9 10 -1 Solution Text Example cont. Step 1 Locate the vertices. To do this, we must identify a2. Based on the standard form of the equation with a horizontal transverse axis, the vertices are a units to the right and to the left of the center. Because a2 = 16, a = 4. This means that the vertices are 4 units to the right and left of the center, (2, 3). Four units to the right of (2, 3) puts one vertex at (6, 3). Four units to the left puts the other vertex at (-2, 3).

  22. 8 7 6 5 4 (-2, 3) (6, 3) 3 (2, 3) 2 1 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8 9 10 -1 Text Example cont. Solution Step 2Draw a rectangle. Because a2 = 16 and b2 = 9, a = 4 and b = 3. The rectangle passes through points that are 4 units to the right and left of the center (the vertices are located here) and 3 units above and below the center. The rectangle is shown using dashed lines.

  23. 8 7 6 5 4 (-2, 3) (6, 3) 3 (2, 3) 2 1 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8 9 10 -1 Text Example cont. Solution Step 3Draw extended diagonals of the rectangle to obtain the asymptotes. We draw dashed lines through the opposite corners of the rectangle to obtain the graph of the asymptotes. The equations of the asymptotes of the hyperbola are y - 3 =  3/4 (x - 2)

  24. 8 7 6 5 4 (-2, 3) (6, 3) 3 (2, 3) 2 1 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8 9 10 -1 Text Example cont. Solution Step 4Draw the two branches of the hyperbola by starting at each vertex and approaching the asymptotes. The hyperbola is shown below. The foci are located c units to the right and left of the center. We find c using c2 = a2 + b2. c2 = 16 + 9 = 25 Because c2 = 25, c = 5. This means that the foci are 5 units to the right and left of the center, (2, 3). Five units to the right of (2, 3) puts one focus at (7, 3). Five units to the left of the center puts the other focus at (-3, 3).

  25. The Parabola

  26. Parabola Directrix Focus Axis of Symmetry Vertex Definition of a Parabola • A Parabola is the set of all points in a plane that are equidistant from a fixed line (the directrix) and a fixed point (the focus) that is not on the line.

  27. The standard form of the equation of a parabola with vertex at the origin is = 4px or x2 = 4py. The graph illustrates that for the equation on the left, the focus is on the x-axis, which is the axis of symmetry. For the equation of the right, the focus is on the y-axis, which is the axis of symmetry. y y 2 = 4px Focus (p, 0) x 2 = 4py Directrix x = -p Focus (p, 0) Vertex x Vertex Directrix y = -p Standard Forms of the Parabola y x

  28. Example • Find the focus and directrix of the parabola given by: Solution: 4p = 16p = 4Focus (0,4) and directrix y=-4

  29. This is 4p. We can find both the focus and the directrix by finding p. 4p = -8 p = -2 The focus, on the y-axis, is at (0, p) and the directrix is given by y = - p. Text Example Find the focus and directrix of the parabola given by x2 = -8y. The graph the parabola. Solution The given equation is in the standard form x2 = 4py, so 4p = -8. x2 = -8y

  30. 5 Directrix: y = 2 4 Vertex (0, 0) 3 2 1 -5 -4 -3 -2 -1 1 2 3 4 5 -1 (-4, -2) (4, -2) -2 -3 -4 -5 Focus (0, -2) Text Example cont. Find the focus and directrix of the parabola given by x2 = -8y. The graph the parabola. Solution Because p < 0, the parabola opens downward. Using this value for p, we obtain Focus: (0, p) = (0, -2) Directrix: y = - p; y = 2. To graph x2 = -8y, we assign y a value that makes the right side a perfect square. If y = -2, then x2 = -8(-2) = 16, so x is 4 and –4. The parabola passes through the points (4, -2) and (-4, -2).

  31. 7 6 Directrix: x = -5 5 4 3 Focus (5, 0) 2 1 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 -1 -2 -3 -4 -5 -6 -7 Text Example cont. Find the standard form of the equation of a parabola with focus (5, 0) and directrix x = -5. Solution The focus is (5, 0). Thus, the focus is on the x-axis. We use the standard form of the equation in which x is not squared, namely y2 = 4px. We need to determine the value of p. Recall that the focus, located at (p, 0), is p units from the vertex, (0, 0). Thus, if the focus is (5, 0), then p = 5. We substitute 5 for p into y2 = 4px to obtain the standard form of the equation of the parabola. The equation is y2 = 4 • 5x or y2 = 20x.

  32. y2 + 2y + 12x – 23 = 0 This is the given equation. y2 + 2y = -12x + 23 Isolate the terms involving y. y2 + 2y + 1 = -12x + 23 + 1 Complete the square by adding the square of half the coefficient of y. Text Example Find the vertex, focus, and directrix of the parabola given by y2 + 2y + 12x – 23 = 0. Then graph the parabola. Solution We convert the given equation to standard form by completing the square on the variable y. We isolate the terms involving y on the left side. (y + 1)2 = -12x + 24

  33. Text Example cont. Solution To express this equation in the standard form (y – k)2 = 4p(x – h), we factor –12 on the right. The standard form of the parabola’s equation is (y + 1)2 = -12(x – 2) We use this form to identify the vertex, (h, k), and the value for p needed to locate the focus and the directrix. (y – (-1))2 = -12(x – 2) The equation is in standard form. We see that h = 2 and k = -1. Thus, the vertex of the parabola is (h, k) = (2, -1). Because 4p = -12, p = -3. Based on the standard form of the equation, the axis of symmetry is horizontal. With a negative value for p and a horizontal axis of symmetry, the parabola opens to the left. We locate the focus and the directrix as follows. Focus: (h + p, k) = (2 + (-3), -1) = (-1, -1) Directrix: x = h – p x = 2 – (-3) = 5 Thus, the focus is (-1, -1) and the directrix is x = 5.

  34. (y + 1)2 = -12(-1 – 2) Substitute –1 for x. (y + 1)2 = 36 Simplify. y + 1 = 6 or y + 1 = -6 Write as two separate equations. y = 5 or y = -7 Solve for y in each equation. Text Example cont. Solution To graph (y + 1)2 = -12(x – 2), we assign x a value that makes the right side of the equation a perfect square. If x = -1, the right side is 36. We will let x = -1 and solve for y to obtain points on the parabola.

  35. Directrix: x = 5 7 6 5 (-1, 5) 4 3 2 1 -5 -4 -3 -2 1 3 4 6 7 -2 -3 Focus (-1, -1) -4 Vertex (2, -1) -5 -6 (-1, -7) -7 Text Example cont. Solution Because we obtained these values of y for x = -1, the parabola passes through the points (-1, 5) and (-1, -7). Passing a smooth curve through the vertex and these two points, we sketch the parabola below.

  36. The Latus Rectum and Graphing Parabolas • The latus rectum of a parabola is a line segment that passes through its focus, is parallel to its directrix, and has its endpoints on the parabola.

  37. Rotation of Axis

  38. Identifying a Conic Section without Completing the Square • A nondegenerate conic section of the form Ax2 + Cy2 + Dx + Ey + F = 0 • in which A and C are not both zero is • a circle if A = C. • a parabola if AC = 0 • an ellipse if A = C and AC > 0, and • a hyperbola if AC < 0.

  39. Solution We use A, the coefficient of x2, and C, the coefficient of y2, to identify each conic section. a. 4x2 – 25y2 – 24x + 250y – 489 = 0 A = 4 C = -25 Text Example • Identify the graph of each of the following nondegenerate conic sections. • 4x2 – 25y2 – 24x + 250y – 489 = 0 • x2 + y2 + 6x – 2y + 6 = 0 • y2 + 12x + 2y – 23 = 0 • 9x2 + 25y2 – 54x + 50y – 119 = 0 AC = 4(-25) = -100 < 0. Thus, the graph is a hyperbola.

  40. Solution b. x2 + y2 + 6x – 2y + 6 = 0 Because A = C, the graph of the equation is a circle. A = 1 C = 1 c. We can write y2 + 12x + 2y – 23 = 0 as 0x2 + y2 + 12x + 2y – 23 = 0. A = 0 C = 1 AC = 0(1) = 0 Because AC = 0, the graph of the equation is a parabola. d. 9x2 + 25y2 – 54x + 50y – 119 = 0 A = 9 C = 25 AC = 9(25) = 225 > 0 Because AC > 0 and A = 0, the graph of the equation is a ellipse. Text Example cont.

  41. Rotation of Axes Formulas • Suppose an xy-coordinate system and an x´y´-coordinate system have the same origin and  is the angle from the positive x-axis to the positive x´-axis. If the coordinates of point P are (x, y) in the xy-system and (x´, y´) in the rotated x´y´-system, then • x = x´cos – y´sin  • y = x´sin  + y´cos .

  42. Amount of Rotation Formula • The general second-degree equation • Ax2 + Bxy + Cy2 + Dx + Ey + F = 0, B = 0 • can be rewritten as an equation in x´ and y´ without an x´y´-term by rotating the axes through angle , where

  43. Writing the Equation of a Rotated Conic in Standard Form • Use the given equation Ax2 + Bxy + Cy2 + Dx + Ey + F=0, B = 0 to find cot 2. • Use the expression for cot 2 to determine , the angle of rotation. • Substitute in the rotation formulas x = x´cos – y´sin andy = x´sin  + y´cos  and simplify. • Substitute the expression for x and y from the rotation formulas in the given equation and simplify. The resulting equation should have no x´y´-term. • Write the equation involving x´ and y´ without in standard form.

  44. Identifying a Conic Section without a Rotation of Axis • A nondegenerate conic section of the form Ax2 + Bxy + Cy2 + Dx + Ey + F = 0, B = 0 is • a parabola if B2 – 4AC = 0, • an ellipse or a circle if B2 – 4AC < 0, and • a hyperbola if B2 – 4AC > 0.

  45. Identify the graph of 11x2 + 103 xy + y2 – 4 = 0. _ Solution We use A, B, and C, to identify the conic section. _ 11x2 + 103 xy + y2 – 4 = 0 A = 11 B = 10 3 C = 1 _ _ B2 – 4AC = (103 )2 – 4(11)(1) = 1003 – 44 = 256 > 0 Text Example Because B2 – 4AC > 0, the graph of the equation is a hyperbola.

  46. Example • Write the equation xy=3 in terms of a rotated x`y`-system if the angle of rotation from the x-axis to the x`-axis is 45º Solution:

  47. Example • Write the equation xy=3 in terms of a rotated x`y`-system if the angle of rotation from the x-axis to the x`-axis is 45º Solution:

  48. Parametric Equations

  49. This is the ball’s horizontal distance, in feet. This is the ball’s vertical height, in feet. Plane Curves and Parametric Equations You throw a ball from a height of 6 feet, with an initial velocity of 90 feet per second and at an angle of 40º with the horizontal. After t seconds, the path of the ball can be described by x = (90 cos 40º)t and y = 6 + (90 cos 40º)t – 16t2. Using these equations, we can calculate the location of the ball at any time t. For example, to determine the location when t = 1 second, substitute 1 for t in each equation: x = (90 cos 40º)(1)  68.9 feet. y = 6 + (90 cos 40º)(1) – 16 (1)2 47.9 feet.

  50. t= 1 sec x= 68.9 ft y= 47.9 ft t= 2 sec x= 137.9 ft y= 57.7 ft t= 3 sec x= 206.8 ft y= 35.6 ft y (feet) 60 40 20 x (feet) 40 80 120 160 200 240 Plane Curves and Parametric Equations This tells us that after 1 second, the ball has traveled a horizontal distance of approximately 68.9 feet, and the height of the ball is approximately 47.9 feet. The figure below displays this information and the results for calculations corresponding to t = 2 seconds and t = 3 seconds.

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